What are the procedures to analyze the similarity of graphs?

Question

This question is taken from a very challenging calculus problems book called the Advanced Problems in Mathematics by Vikas Gupta

And I have absolutely no clue on how to approach questions of the kind. My initial thought since y is composite function was to check for a graph similar to the outermost function which is that of tan. Now option D looked compelling. But I was incorrect.

Then I decided to substitute values like say $x=\pi/4$ but that would yield $\tan(1/{\sqrt{2}})$ which is hard to compute.

So please do advise me on how to approach this question efficiently. Thanks

Answer 1

Let's quickly eliminate two cases: a and d. Take $x=\pi/2$. Then $\sin x =1$. I don't know the exact value of $\tan 1$, but I know that $1>\pi/4$ and that $\tan$ is increasing between $0$ and $\pi/2$. So $\tan 1>1$. Therefore no option a. Similarly, $\sin x$ has values between $-1$ and $1$. The $\tan$ function is continuous between $\pi/2$ and $\pi/2$, so it is not diverging in the $[-1,1]$ interval. Therefore d option is not right.

So what is left is to distinguish between b and c. One simple way is to calculate the derivative at $\pi/2$. In one case it is zero, in the other case you have left and right limits that are different. $$\frac{d\tan(\sin x)}{dx}=\frac{\cos x}{\cos^2(\sin x)}$$ $\cos \frac\pi 2=0$, $\sin\frac\pi 2=1$,$\cos 1\ne 0$. Therefore the derivative is $0$. This is case b. Note that this also eliminates case d.

Answer 2

First, $\sin (x) \in [-1,1]$ and for all these values, $\tan$ is defined nicely.

Also, the function $\tan(\sin(x))$ is differentiable at all points.

That removes options (c) and (d) from consideration.

Now look at option (a) and (b). You only have to figure out if $\tan(\sin(\pi/2)) > 1$ or not; which you can do easily. Observe that $\tan(1) > \tan (\pi/4)$ hence $\tan(1) > 1$ .

Thus the answer is option (b).